91Ó°ÊÓ

Acceleration is a crucial concept in kinematics. It measures how quickly an object's velocity changes with time. In this exercise, the acceleration of the bus is described by the linear function \(a_{x}(t) = \alpha t\), with \(\alpha = 1.2 \, \mathrm{m/s}^3\). This equation indicates that acceleration increases as time passes, forming a ramp or a line through the origin.
Over time, if the slope \(\alpha\) is positive, the velocity increases; if \(\alpha\) is negative, it decreases. Here, since \(\alpha\) is positive, the bus accelerates as time unfolds. This constant increase leads to an interesting scenario where both velocity and position must be calculated through integration.

The velocity function tells us how fast the bus is moving at any given time. It is derived from the acceleration function by performing integration. Given that the acceleration is \(a_x(t) = \alpha t\), integrating this with respect to time gives us:\[v_x(t) = \int a_x(t) \, dt = \frac{\alpha}{2} t^2 + C\]Where \(C\) is the constant of integration which can be found using initial conditions. In our problem, we know that the velocity at \(t = 1.0 \, \mathrm{s}\) is \(5.0 \, \mathrm{m/s}\), helping us solve for \(C\):
\[5.0 = 0.6 \cdot 1^2 + C\]Thus, \(C = 4.4\). This adjustment gives us the formula:
\[v_x(t) = 0.6 t^2 + 4.4\]
By plugging in \(t = 2.0 \, \mathrm{s}\), one can compute the corresponding velocity, which turns out to be \(6.8 \, \mathrm{m/s}\). This illustrates how the rate of speed grows due to continual acceleration.

The position of the bus is determined by a position function, which is sourced from integrating the velocity function. This function accounts for the distance traveled by the bus. Given the velocity function \(v_x(t) = 0.6 t^2 + 4.4\), the position function is determined as:\[x(t) = \int v_x(t) \, dt = \frac{0.6}{3} t^3 + 4.4 t + C'\]Here, \(C'\) is another integration constant. We learn from the problem that the initial position at \(t = 1.0 \, \mathrm{s}\) is \(6.0 \, \mathrm{m}\). So:
\[6.0 = 0.2 + 4.4 \times 1 + C'\]This results in \(C' = 1.4\), giving us the final position function:
\[x(t) = 0.2 t^3 + 4.4 t + 1.4\]
By substituting \(t = 2.0 \, \mathrm{s}\) into this equation, we find that the position at this time is \(11.8 \, \mathrm{m}\). This demonstrates how the position expands as a cube function due to the cubic term from the continually increasing velocity.

Integration is a powerful tool in physics, particularly in kinematics, for finding functions like velocity and position from known acceleration functions. It involves a process of calculating the total accumulation of a quantity, like speed or distance, over time based on known rates of change. In our exercise:

• To find the velocity function, we integrated the acceleration function \(a_x(t) = \alpha t\), resulting in a quadratic model of velocity.
• Next, integrating the derived velocity yielded the position function, showing how the cumulative effects of acceleration and velocity determine positional shifts over time.

Key Points:

• Each integral introduces a constant, \(C\), accounting for initial conditions like initial velocity or initial position.
• Understanding integration helps describe how objects move, providing deeper insight into real-world motion than simply knowing instantaneous changes like acceleration.

This knowledge anchors much of classical mechanics and enhances problem-solving skills for real-world applications.